For the following, assume that all the given angles are in simplest form, so that if A is in QIV you may assume that 270° < A < 360°. If sin A = with A in QII, find the following. cos
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![### Trigonometric Functions and Angle Determination
#### Problem Statement
For the following, assume that all the given angles are in simplest form, so that if \( A \) is in Quadrant IV (QIV), you may assume that \(270^\circ < A < 360^\circ\).
Given:
\[ \sin A = \frac{3}{5} \]
with \( A \) in Quadrant II (QII), find the following:
\[ \cos \frac{A}{2} \]
#### Solution
The problem provides us with the sine value of angle \( A \) and specifies that angle \( A \) is in the second quadrant (QII). We are asked to find the value of cosine for half of angle \( A \).
A diagram is presented that incorrectly calculates the value as follows:
\[ \cos \frac{A}{2} = -\sqrt{\frac{9}{2} \left(\frac{5}{2}\right)} \]
This solution is incorrect, as indicated by a red X mark beside the expression. Therefore, it is important to use the correct trigonometric formulas and consider the quadrant details to find the correct value.
#### Correct Approach
1. **Trigonometric Identity**:
Using the half-angle formula for cosine:
\[
\cos \frac{A}{2} = \pm \sqrt{\frac{1 + \cos A}{2}}
\]
Because \( A \) is in the second quadrant, \( \cos A \) will be negative. We first need to determine \( \cos A \).
2. **Determining \( \cos A \)**:
\[
\sin^2 A + \cos^2 A = 1
\]
\[
\left(\frac{3}{5}\right)^2 + \cos^2 A = 1
\]
\[
\frac{9}{25} + \cos^2 A = 1
\]
\[
\cos^2 A = 1 - \frac{9}{25}
\]
\[
\cos^2 A = \frac{16}{25}
\]
Since \( A \) is in QII, \( \cos A \) is negative:
\[
\cos A = -\frac{4}{5}
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